Question

PLEASE HELP!!! WILL GIVE BRAINLIEST!!!

Answer 1

Answer: (2,2) and (-5,16)

Step-by-step explanation:

Here we have both Line (Linear Function) and Parabola (Quadratic Function)

So I am gonna write these equations here,

$y=x^2+x-4\\y=-2x+6$

The first equation has Parabola graph (Since it's second degree.)

and the second equation has line graph.

To find the intersection, you have to substitute either -2x+6 in first equation (Quadratic) or x^2+x-4 in second equation (Linear)

For me, I am going to substitute x^2+x-4 in y=-2x+6.

$x^2+x-4=-2x+6$

Now solve the equation and find the value of x.

Since it's Quadratic Equation (Because there's x^2) I'd move -2x+6 to the left side.

$x^2+x-4+2x-6=0$ Finish things here (Subtract and Addition)

$x^2+3x-10=0$ What two numbers multiply to 10? Find the factors of 10, that are [1 and 10] and [2 and 5]

Now think about it, do you think that if 1 and 10 subtract or even addition, do you think that it'd be 3? No, of course not.

So 2 and 5 is right.

$(x-2)(x+5)=0$ (5-2 = 3) and (5*(-2) = -10)

Then we get both x, $x=2,-5$

However, this is not it. You have to substitute both x in Linear Equation.

Substitute x = 2 in y=-2x+6

$y=-2(2)+6\\y=-4+6\\y=2$

Order = $(2,2)$

Then substitute x = -5 in y=-2x+6

$y=-2(-5)+6\\y=10+6\\y=16$

Order = $(-5,16)$

So the intersections are both (2,2) and (-5,16) as shown in graph below.